Question

Найдите значение выражения. 20 баллов!
$\sqrt{2}- 2\sqrt{2}sin^2\frac{15\pi }{8}$

Answer 1

Ответ:  1

Объяснение:

$2sin^2x=1-cos2x\\\\2sin^2{\frac{15\pi}{8} }=1-cos{\frac{30\pi}{8} }=1-cos{\frac{15\pi}{4} }=1-cos{\frac{15-8\pi}{4}}=1-\frac{\sqrt{2} }{2} \\\\\sqrt{2}-\sqrt{2}\cdot (1-\frac{\sqrt{2} }{2})=\sqrt{2}-\sqrt{2}+\frac{\sqrt{2}\sqrt{2} }{2}=1$

Answer 2

$\sqrt{2}-2\sqrt{2}\sin^2\dfrac{15\pi}{8}=\sqrt{2}-2\sqrt{2}\cdot \dfrac{1-\cos (2\cdot \frac{15\pi}{8})}{2}=\sqrt{2}\cos \dfrac{15\pi}{4}=\\ \\ \\ =\sqrt{2}\cos \left(4\pi-\dfrac{\pi}{4}\right)=\sqrt{2}\cos \dfrac{\pi}{4}=\sqrt{2}\cdot \dfrac{1}{\sqrt{2}}=1$